Question

a car accelerates uniformly from 36 km per hour to 72 km per hour in 10 seconds.
1)calculate the acceleration? 2)calculate distance covered by the car in that time?

Answer 1

convert 36 km/h and 72km/h into m/s u will get 10 m/s for 36km/h and 20m/s for 72 km/h.
initial velocity =10m/s
final velocity= 20m/s
time =10 sec
acceleration = (v-u)÷t
=(20-10)÷10
=1 m/s^2
distance= ut +1/2at^2
=10×10+1/2×1×10×10
=100+ 50
=150m

Answer 2

Concept:

• One dimensional motion
• Kinematic equations
• Acceleration, velocity

Given:

• Initial speed u= 36km/h = 10m/s
• Final speed v = 72km/h = 20m/s
• Time in which the speed is changed = 10s

Find:

• The acceleration of the car
• The distance covered by the car the 10 seconds

Solution:

v = u +at

20 = 10 + 10a

10 = 10a

a = 1m/s^2

s = ut +1/2at^2

s = 10(10) +1/2 (1) (10)(10)

s = 100+ 50

s = 150m

The acceleration of the car is 1m/s².

The distance covered by the car during the acceleration is 150m.

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